How can I show that $(I_1 × · · · × I_n) \setminus \{a\}$ is open and connected for any $a ∈ \mathbb{R}^n$ (with $n \geq 2$)? Let $I_1, . . . , I_n$ be any open intervals in $\mathbb{R}$ for $n ≥ 2$. How can I show that $(I_1 × · · · × I_n) \setminus \{a\}$ is open and connected for any $a ∈ \mathbb{R}^n$?
 A: The set $I_1\times I_2\times\dots\times I_n$ is open by definition of product topology. The set $\{a\}$ is closed because $\mathbb{R}^n$ is Hausdorff. So
$$
(I_1\times I_2\times\dots\times I_n)\setminus\{a\}=
(I_1\times I_2\times\dots\times I_n)\cap(\mathbb{R}^n\setminus\{a\})
$$
is open because it's the intersection of two open sets.
If $a\notin I_1\times I_2\times\dots\times I_n$, then the set is connected, being the product of connected spaces (it's actually convex and open, so connected by arcs).
If $a\in I_1\times I_2\times\dots\times I_n$, then the set is still connected by arcs. Take two points $b$ and $c$ in $(I_1\times I_2\times\dots\times I_n)\setminus\{a\}$. If the segment joining $b$ and $c$ doesn't pass through $a$ we are done. If it passes through $a$, then consider $\varepsilon>0$ such that the $n$-sphere with center $a$ and radius $\varepsilon$ is contained in $I_1\times I_2\times\dots\times I_n$.
Pick the point $b'$ where the segment $ba$ meets the $n$-sphere and fix similarly $c'$; consider and a point $d$ inside the $n$-sphere such that neither the segment $b'd$ nor the segment $dc'$ pass through $a$. Then the polygonal $bb'dc'c$ is a path joining $b$ and $c$ contained in $(I_1\times I_2\times\dots\times I_n)\setminus\{a\}$.
A: Use the fact that $\mathbb{R}^n$ is a metric space (which is Hausdorff). If the induced topology on  $(I_1 × · · · × I_n)$ is Hausdorff then $\{a\}$ is closed, but in $(I_1 × · · · × I_n)$ and so $(I_1 × · · · × I_n) \setminus \{a\}$ is open, but in $(I_1 × · · · × I_n)$.
